Reducing variability

• The paired data sets in this chapter have one pair of quantitative response values for each observational unit, allowing for a built-in comparison.
• Studies with paired data remove individual variability by looking at the difference score for each individual.
• Reducing variability in data improves inferences:
  • Narrower confidence intervals
  • Smaller p-values when the null hypothesis is false (more power)

Hypotheses

• \(H_0: \mu_\text{diff} = 0\), the average tread wear is the same for the two tire brands.
• \(H_A: \mu_\text{diff} \neq 0\), the average tread wear is different across the two tire brands.

Observed Statistic: \(\bar{x}_\text{diff} = -0.002\)

Preview

    JJ bicycle
1  118     118
2  146     124
3  134      92
4   94      80
5  146     111
6  114     112
7  132     124
8  118     124
9  116      97
10 101      82
11 112     143
12  99      97
13 112      88
14 118     123
15  88      83
16  70      73
17 106      87
18 108     137
19 124      98
20 143      96
21  92      70
22 131     100

Group Activity

1. Go to the Matched Pairs applet, and select the JJvsBicycle data from the dropdown menu. Discuss the following:
   • In the top graph, what do the red dots, blue dots, and gray lines represent?
   • In the bottom graph, what do the dots represent?
2. Check the box for Randomize and then click on the Randomize button.
   • What do the coin flips determine?
   • How is the dot on the null distribution calculated?
3. Perform 1000 (or more) randomizations, and record a p-value for the above test. In the context of the problem (exercise), record a sentence saying what this p-value tells us.

Now do it wrong

• If you mistakenly use the two-mean test, you will lose power.
• See the Multiple Means applet, and try the two-mean test on this data.

21.3.1 Observed data

ggplot(textbooks, aes(x = price_diff)) + geom_histogram(binwidth = 5)

21.3.2 Variability of the statistic

textbooks %>%
  summarize(n = n(),
            meanDiff = mean(price_diff),
            sdDiff = sd(price_diff))

# A tibble: 1 × 3
      n meanDiff sdDiff
  <int>    <dbl>  <dbl>
1    68     3.58   13.4

Hypotheses

• \(H_0: \mu_\text{diff} = 0\), there is no difference in the average textbook price between the bookstore and Amazon
• \(H_A: \mu_\text{diff} \neq 0\), there is a difference in the average price.

21.3.3 Observed statistic vs. null statistics

The mathematical model is very similar to the one-mean case:

\[ T = \frac{\bar{x}_\text{diff} - 0}{s_\text{diff}/\sqrt{n}} \]

The conditions are similar: Independence, large samples, or normality of the differences.

Confidence interval for the mean difference

A confidence interval for \(\mu_\text{diff}\) is:

\[ \bar{x}_\text{diff} \pm t^\star_\text{df} \times \frac{s_\text{diff}}{\sqrt{n}} \]

As before, \(df = n-1\). Use qt() to get \(t^\star\).

Have R compute the differences

t.test(textbooks$bookstore_new, textbooks$amazon_new, paired = TRUE)

    Paired t-test

data:  textbooks$bookstore_new and textbooks$amazon_new
t = 2.2012, df = 67, p-value = 0.03117
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 0.3340641 6.8324064
sample estimates:
mean difference 
       3.583235 

Just do a one-sample test on the differences

t.test(textbooks$price_diff)

    One Sample t-test

data:  textbooks$price_diff
t = 2.2012, df = 67, p-value = 0.03117
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 0.3340641 6.8324064
sample estimates:
mean of x 
 3.583235